Beam Theory Explained - How Spars Work

HomeBuiltAirplanes.com

Help Support HomeBuiltAirplanes.com:

Status
Not open for further replies.

PiperCruisin

Well-Known Member
Joined
Jan 17, 2017
Messages
146
Location
Idaho
I don't disagree and think it is a great thread. There is a lot of knowledge to be shared on beams. Everyone should realize there is no substitution for diving into some mechanics of materials and get into the math if one wants to pursue building an aircraft structure.

Beams aside, the trick is often how to shear the load into and out of the caps. How fast can you do it without a delamination of the joint or overloading one of the fasteners causing it to unzip because the load was not distributed. Without a proper introduction of the load, the beam will not realize its full potential.

I'll see if I can dig up some photos and start a thread showing some examples of joints.
 

Turd Ferguson

Well-Known Member
Joined
Mar 13, 2008
Messages
4,944
Location
Upper midwest in a house
This should be pinned or put in a library topic where it would be easy to find.

What's with all the duplicate posts in this thread? Is it the software or is it just me?
 

geosnooker2000

Well-Known Member
Joined
Mar 30, 2019
Messages
108
Location
Somerville, TN
So, for the dummies in class, I think it would be instructive for someone who has done one before, to run through an example calculation on an actual spar in an actual wing, showing how the loading diagram was arrived at. For my part, I will jump out and show my ignorance by guessing for y'all's amusement how a spar is loaded, just to demonstrate what some of us need help in understanding.

Keep in mind, I have a masters in Architecture, which means I had 3 semesters of structures, so I am coming at this with calculating beams mostly intended to withstand only forces in the Z axis (with the exception of steel columns - something about radius of gyration that I forgot as soon as I passed the class). We DID get into truss design, so I "had" a good grasp of vector addition, but boy do I need a refresher...

Example. a low-wing aluminum airplane with a span of 32'. Fuselage is 4' wide. Leaves us with a 14' cantilevered beam (Wing Spar). Assuming a tapered wing design, the loads in the positive G category would be a gradual increase as they reach the attached end. There would not be a point load on the outboard end of the beam diagram, even though I imagine picking up a model airplane by the tips of the wings to see if it is within acceptable CG - That is not how the air (wind) will pick up the plane. it will pick it up at every instance along the underside of the spar.

So my first thought was, yeah, it's just a unequal uniform load consisting of half the weight of the plane times however many G-forces you are designing for (times 1.5 safety margin). Then I thought, no... the wing ribs will intersect and attach at, for example, every 16" o.c. Maybe it should be a series of point loads, each point load conveying the area each rib caries. In a building, I would not make the distinction, as wood and steel are far to stiff to worry about buckling @ 16" o.c. braced, but we are talking about aluminum here.

Well, that should be enough to start a beat-down-fest of my ignorance. Have I got ANY of that right?
 

proppastie

Well-Known Member
Log Member
Joined
Feb 19, 2012
Messages
4,175
Location
NJ
https://www.homebuiltairplanes.com/forums/threads/wing-spar-sizing-design.27939/#post-391463

The search function is a big help too. I like the EAA metal design spread sheet in the thread. Also "The Glider" by Frati also has the simplified loading I used. Others will explain that it is elliptical loading but the platform loading is close enough, more conservative, and easier to understand. I have been told it is like calculating the live load on a roof, but that I would not know.
 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
Hmm, well, I could do one example of the air loading, shear, bending moment, curvature, and deflection. I could even do the exact solution and a numerical integration to show how close they are.

To get into geosnooker2000's questions:

  • The air load falls off toward the tips - air leaks around the end. Even a rectangular wing tends to look like it has an elliptical distribution of spanwise load. That complicates things a bit. Then if the wing is tapered or has washout either by twisting the wiing or by using a more highly cambered foil towards the tip, things get more complicated;
  • The airload for the wing does carry all the way to the centerline, even when a bulky fusealge is in there. The pressures developed on the wing can not just disappear at the wing, so the pressure distribution looks elliptical all the way to the centerline, and we get our airloads from that;
  • The airload summed up on the entire span is actually equal to the weight of the airplane times the g's plus whatever download the tail is making to neutral out the airplane pitching moments. If the tail is making positive lift, it does then reduce the lift needed from the wing, but you do not find much of that except in canard ships;
  • In a fabric covered wing, the ribs do collect the airloads and drop that load in at the rib mounts. If you have a plywood D-tube, then the skin ends up distributing the some of the lift along the spar, with a reduced point load at each rib mount. The errors from either going smooth with an exact analytical approach or doing the wing piece-wise at the rib spacing are small, as are the errors from adding fidelity or not from how the load is collected and applied to the spar. Get into composites and we tend to either have one continuous rib (massive foam core) or very few ribs, and the lifting load is nicely spread out onto the spar;
  • In all skinning approaches, skin strength and airloads end up setting how far apart we can allow ribs to get... A whole 'nother topic.

I will put together some sort of an example, but it may be a few days out.

Billski
 

Dave Hodges

Well-Known Member
Joined
Jun 29, 2014
Messages
81
Location
Reidsville, GA
So, for the dummies in class, I think it would be instructive for someone who has done one before, to run through an example calculation on an actual spar in an actual wing, showing how the loading diagram was arrived at. For my part, I will jump out and show my ignorance by guessing for y'all's amusement how a spar is loaded, just to demonstrate what some of us need help in understanding.

Keep in mind, I have a masters in Architecture, which means I had 3 semesters of structures, so I am coming at this with calculating beams mostly intended to withstand only forces in the Z axis (with the exception of steel columns - something about radius of gyration that I forgot as soon as I passed the class). We DID get into truss design, so I "had" a good grasp of vector addition, but boy do I need a refresher...

Example. a low-wing aluminum airplane with a span of 32'. Fuselage is 4' wide. Leaves us with a 14' cantilevered beam (Wing Spar). Assuming a tapered wing design, the loads in the positive G category would be a gradual increase as they reach the attached end. There would not be a point load on the outboard end of the beam diagram, even though I imagine picking up a model airplane by the tips of the wings to see if it is within acceptable CG - That is not how the air (wind) will pick up the plane. it will pick it up at every instance along the underside of the spar.

So my first thought was, yeah, it's just a unequal uniform load consisting of half the weight of the plane times however many G-forces you are designing for (times 1.5 safety margin). Then I thought, no... the wing ribs will intersect and attach at, for example, every 16" o.c. Maybe it should be a series of point loads, each point load conveying the area each rib caries. In a building, I would not make the distinction, as wood and steel are far to stiff to worry about buckling @ 16" o.c. braced, but we are talking about aluminum here.

Well, that should be enough to start a beat-down-fest of my ignorance. Have I got ANY of that right?
If you want to brush up on beam design, I hit on it in one of my calculus videos. I forget which one. But if you want to get started, here you go:
 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
Ok, I ran through an example with elliptical lift distribution, tailored spars, and numerical integration, then thought maybe I ought to give a simpler example first. I still ran a beam supported in the middle but with the load uniformly distributed over the span, figured out shear and bending moment, designed a straight beam section to support all that load, and then figured angles and vertical deflection.

Now if you still want to see all that other stuff, I can proof read it and make it decent and set that up on here too, but I will not bother if you do not ask for it. Grin.

PS - I caught a couple errors, and they are now fixed. If you have already loaded a copy of the attached files, please discard them and copy a fresh file.

Billski
 

Attachments

Last edited:

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
OK, since I am getting some likes and downloads, here is the text and spreadsheet for an elliptically distributed lift and a first approximation on a tailored spar.

I hope that this helps with understanding on both how the calcs work and how to program efficiently in Excel.

Vagaries in Explorer... you guys should be able to see the elliptical lift distribution files now.

Billski
 

Attachments

Last edited:

BBerson

Well-Known Member
HBA Supporter
Joined
Dec 16, 2007
Messages
12,796
Location
Port Townsend WA
I don't use Excel. Using my hand calculations and a sample procedure on page 411 of Aircraft Structures, I find that .020 2024 web is enough for 500 pound gross ultralight. Seems too thin, but 500 is a quarter of 2000 pounds and .020 is a quarter of .080.
 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
I don't use Excel. Using my hand calculations and a sample procedure on page 411 of Aircraft Structures, I find that .020 2024 web is enough for 500 pound gross ultralight. Seems too thin, but 500 is a quarter of 2000 pounds and .020 is a quarter of .080.
"Other researchers have come to similar conclusions" translates to "somebody else thinks so too". hehehe...

We still have to prevent buckling, which in both of these webs will likely be an issue too. There are probably stiffeners in the future for both of these spars unless there are a lot of ribs...

Billski
 

BBerson

Well-Known Member
HBA Supporter
Joined
Dec 16, 2007
Messages
12,796
Location
Port Townsend WA
My example calculation was with stiffeners at 5" spacing. Which gave a factor or safety of 2.4. Don't know if that is non- buckling or limited elastic buckling.
Thanks for the Bruhn chapter.
 

geosnooker2000

Well-Known Member
Joined
Mar 30, 2019
Messages
108
Location
Somerville, TN
Check out the thickness on these main spars at the business end. What do you suppose that thickness is? In total, and individual sheets. Are those just gussets that penetrate into the first web? And 4 bolts at the bottoms and 4 bolts at the tops? That's all you need to stave off roughly 6000 lbs of shear? That's 6000 lbs / 8 bolts = 750 lbs per bolt... I guess that seems reasonable. I would have just thought those connections would go all the way to the mid-point of the fuselage (2' long) under the seats and connect with a gusset that is the whole width of the cabin (4').

"quick build" wings for a Sling 4

 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
Something to know about wing spars. At the root, shear is the accumulated lift of that wing, while bending is the lift times the distance to about 40% out on the wing. Moments are huge compared to shear... That width you see at the top and bottom are probably the width of the caps need to carry the bending load converted into tensile load (in one cap) and compressive load (in the other cap). The lateral loads through the bolts will be an order of magnitude larger than the vertical loads. The shear webs are probably that thick at the joints simply to help collect the loads to the bolted joints.

Metal wings are frequently done this way. We can collect the loads and carry them compactly - that is in small volume - in metal structures. In composite wings hardpoints tend to be really bulky, so we tend to overlap spars to get the forces down and then use only a couple bolts or large diameter thin wall pins.

There are what I count as four bolts top and four more bottom. The bolts are in shear twice each, once on each side of the spar, so there is 16 times the area of one bolt. It is common to design the joints with bolts or rivets to a FOS of 4 while the rest of the aluminum structure uses FOS of 1.5. And they appear to be AN6 bolts, which are 3/8 diameter and 125,000 psi steel. That works out to about 8000 pounds of shear strength per shear surface... The aluminum is probably a 33,000 or 37,000 psi material, and so it needs more bearing area than the bolts do... Shigley is the book to go to for an intro to bolted joint design.

The Sling aircraft work fine. You want to get freaked out, look at a C172 in the maintenance shop with the root covers off. One bolt at the main spar, one at the drag spar, and two little AN bolts at each end of the lift strut. That's all it needs and they work fine... But each of these designs was engineered carefully.

Billski
 
Last edited:

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,295
Location
Saline Michigan
I just put the Sling 4's specifics into the spreadsheet I supplied, added in a calc for bolt size needed and that stuff all looks about right...
 
Last edited:

geosnooker2000

Well-Known Member
Joined
Mar 30, 2019
Messages
108
Location
Somerville, TN
Note that, when I said 6000 lbs of shear, I got that from total w @ a cantilever beam support end= shear
1000 lbs (half the max gross weight) x 4Gs = 4000 lbs x 1.5 factor = 6000 lbs.
 
2
Status
Not open for further replies.
Group Builder
Top